Resultant force on curved surface BCDEF

The resultant force on a curved surface has a horizontal component (force on the vertical projection) and a vertical component (weight of the liquid vertically above the surface up to the equivalent free surface). The figure is not fully readable, so the geometry below is assumed.

Assumed geometry (water on the concave side, length $L = 10$ m, gauge at level of B reads $-19$ kN/m²):

   B |  gauge level, p = -19 kPa
     | BC = 2 m (vertical wall)
   C |
      \  CD, DE = two quarter circles,
       ) D  R = 2 m (semicircle C-D-E)
      /
   E |________ F   EF = 3 m (horizontal)

Vertical extent of B to E = 2 + 2 + 2 = 6 m, and EF lies at 6 m below B.

Pressure at B and equivalent free surface

$$\frac{p_B}{\gamma} = \frac{-19}{9.81} = -1.937\ \text{m}$$

The equivalent free surface is 1.937 m below B. Pressure at depth $z$ below B is $p = -19 + 9.81z$ kPa.

Horizontal component (vertical projection BE, height 6 m):

$$\begin{aligned} F_H &= L\int_0^6 (-19 + 9.81z)\,dz = L\left(-19\times 6 + 9.81\times\frac{6^2}{2}\right)\\ &= 10(-114 + 176.58) = 625.8\ \text{kN} \end{aligned}$$

Vertical component

• On the semicircle CDE: the upper quarter is pushed up and the lower quarter is pushed down by the liquid. The net force equals the weight of liquid in the half-disc enclosed between the two arcs:

$$F_{V1} = \gamma\left(\frac{\pi R^2}{2}\right)L = 9.81\times 6.283\times 10 = 616.4\ \text{kN (downward)}$$

• On the flat EF at 6 m depth: $p = -19 + 9.81\times 6 = 39.86$ kPa

$$F_{V2} = 39.86\times 3\times 10 = 1195.8\ \text{kN (downward)}$$ $$F_V = 616.4 + 1195.8 = 1812.2\ \text{kN}$$

Resultant

$$\begin{aligned} F &= \sqrt{625.8^2 + 1812.2^2} = 1917.2\ \text{kN}\\ \theta &= \tan^{-1}\frac{1812.2}{625.8} = 70.9^\circ \text{ below the horizontal} \end{aligned}$$

Answer: $F \approx 1917$ kN (about 1.92 MN), inclined at $70.9^\circ$ to the horizontal, with $F_H = 625.8$ kN and $F_V = 1812.2$ kN. If the actual figure differs, use the same method: $F_H = \int p\,dA_{proj}$ and $F_V$ = weight of liquid up to the equivalent free surface.